At elevated temperatures, SbCl5 gas decomposes into SbCl3 gas and Cl2 gas as shown by the following equation: SbCl5(g)<=> SbCl3(g) + Cl2(g)
(a) An 59.8 gram sample of SbCl5 (molecular weight 299.0) is placed in an evacuated 15.0 liter container at 182°C.
1. What is the concentration in moles per liter of SbCl5 in the container before any decomposition occurs?
2. What is the pressure in atmospheres of SbCl5 in the container before any decomposition occurs?
b)If the SbCl5 is 29.2 percent decomposed when equilibrium is
established at 182°C, calculate the values for equilibrium constants Kp and Kc, for this decomposition reaction.
Kc=?
Kp=?
c.In order to produce some SbCl5, a 1.00 mole sample of SbCl3 is first
placed in an empty 2.00 liter container maintained at a temperature
different from 182ºC. At this temperature, Kc, equals 0.117. How many moles of Cl2 must be added to this container to reduce the number of
moles of SbCl3 to 0.500 mole at equilibrium?
To solve these problems, we'll need to use the given information and apply the principles of equilibrium and stoichiometry. Let's solve each part step-by-step:
(a) What is the concentration in moles per liter of SbCl5 in the container before any decomposition occurs?
Step 1: Calculate the number of moles of SbCl5 present in the sample:
Moles of SbCl5 = mass / molar mass
Moles of SbCl5 = 59.8 g / 299.0 g/mol = 0.20 mol
Step 2: Calculate the concentration in moles per liter:
Concentration = moles / volume
Concentration = 0.20 mol / 15.0 L = 0.013 mol/L
Answer: The concentration of SbCl5 in the container is 0.013 mol/L before any decomposition occurs.
(b) What is the pressure in atmospheres of SbCl5 in the container before any decomposition occurs?
Step 1: Determine the number of moles of SbCl5:
Moles of SbCl5 = mass / molar mass = 59.8 g / 299.0 g/mol = 0.20 mol
Step 2: Calculate the partial pressure of SbCl5 using the ideal gas law:
PV = nRT
P = nRT / V
P = (0.20 mol)(0.0821 atm·L/mol·K)(455.15 K) / 15.0 L
Answer: The pressure of SbCl5 in the container before any decomposition occurs is approximately 1.21 atm.
(c) If the SbCl5 is 29.2 percent decomposed when equilibrium is established at 182°C, calculate the values for equilibrium constants Kp and Kc, for this decomposition reaction.
Step 1: Calculate the number of moles of SbCl5 that have decomposed:
Moles of SbCl5 decomposed = 29.2% of 0.20 mol
Moles of SbCl5 decomposed = 0.292 * 0.20 mol = 0.0584 mol
Step 2: Calculate the moles of SbCl5 and Cl2 present at equilibrium:
Moles of SbCl5 at equilibrium = 0.20 mol - 0.0584 mol = 0.1416 mol
Moles of Cl2 at equilibrium = 0.0584 mol
Step 3: Calculate the equilibrium concentrations:
Concentration of SbCl5 (C) = moles / volume = 0.1416 mol / 15.0 L
Concentration of Cl2 (C) = moles / volume = 0.0584 mol / 15.0 L
Step 4: Calculate the equilibrium constants:
Kc = ([C(SbCl3)][C(Cl2)]) / ([C(SbCl5)])
Kp = (P(SbCl3) * P(Cl2)) / (P(SbCl5))
Answer: The equilibrium constant Kc is calculated using the equilibrium concentrations of SbCl3, Cl2, and SbCl5, while Kp is calculated using their partial pressures.
(c) In order to produce some SbCl5, a 1.00 mole sample of SbCl3 is first placed in an empty 2.00 liter container maintained at a temperature different from 182ºC. At this temperature, Kc equals 0.117. How many moles of Cl2 must be added to this container to reduce the number of moles of SbCl3 to 0.500 mole at equilibrium?
Step 1: Write the balanced equation for the reaction:
SbCl3(g) + Cl2(g) ⇌ SbCl5(g)
Step 2: Set up the equilibrium expression and substitute the given equilibrium constant and initial moles of SbCl3 and volume:
Kc = ([C(SbCl5)][C(Cl2)]) / ([C(SbCl3)])
0.117 = ([0.500 mol / (2.00 L + x)][x]) / ([1.00 mol / (2.00 L + x)])
Step 3: Solve for x, the moles of Cl2 to be added:
Rearrange the equation and solve for x:
0.117 * [1.00 mol / (2.00 L + x)] = [0.500 mol / (2.00 L + x)] * x
Step 4: Solve the equation for x.
Answer: Solve the equation to find the value of x, which will be the moles of Cl2 that must be added to the container to reduce the number of moles of SbCl3 to 0.500 mole at equilibrium.
To solve these problems, we will use the principles of stoichiometry and equilibrium expressions. Let's break down each question step by step:
(a) What is the concentration in moles per liter of SbCl5 in the container before any decomposition occurs?
To find the concentration of SbCl5 in moles per liter, we need to know the number of moles of SbCl5 and the volume of the container. Given that we have a 59.8 gram sample of SbCl5 and its molecular weight is 299.0 g/mol, we can calculate the number of moles using the formula:
Moles of SbCl5 = Mass of SbCl5 / Molecular weight of SbCl5
Moles of SbCl5 = 59.8 g / 299.0 g/mol = 0.200 mol
Since we know the volume of the container is 15.0 liters, we can now calculate the concentration of SbCl5 in moles per liter:
Concentration of SbCl5 = Moles of SbCl5 / Volume of container
Concentration of SbCl5 = 0.200 mol / 15.0 L = 0.0133 mol/L
Therefore, the concentration of SbCl5 in the container before decomposition occurs is 0.0133 moles per liter.
(b) What is the pressure in atmospheres of SbCl5 in the container, before any decomposition occurs?
To calculate the pressure of SbCl5, we can use the ideal gas law:
PV = nRT
Where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin. First, we need to convert the temperature from Celsius to Kelvin:
T(K) = T(°C) + 273.15
T(K) = 182°C + 273.15 = 455.15 K
Given that the volume of the container is 15.0 liters and the number of moles of SbCl5 is 0.200 mol (from part (a)), and the ideal gas constant is typically 0.0821 L·atm/mol·K, we can calculate the pressure:
P = (nRT) / V
P = (0.200 mol)(0.0821 L·atm/mol·K)(455.15 K) / 15.0 L
P = 5.977 atm
Therefore, the pressure of SbCl5 in the container before decomposition occurs is 5.977 atmospheres.
(b) If the SbCl5 is 29.2 percent decomposed when equilibrium is established at 182°C, calculate the values for equilibrium constants Kp and Kc for this decomposition reaction.
To calculate the equilibrium constants Kp and Kc, we need to know the initial and equilibrium concentrations or partial pressures of the reactants and products. One way to approach this problem is to assume a total initial pressure of "P" and calculate the equilibrium pressures using the given percent decomposition.
Let x represent the decrease in pressure of SbCl5 due to the decomposition:
Pressure of SbCl5 at equilibrium = Initial pressure of SbCl5 - x
Pressure of SbCl3 at equilibrium = Initial pressure of SbCl3 + x
Pressure of Cl2 at equilibrium = Initial pressure of Cl2 + x
Given that the SbCl5 is 29.2% decomposed, x is equal to 29.2% of the initial pressure of SbCl5:
x = 0.292 * Initial pressure of SbCl5
To calculate the equilibrium pressures, we can subtract x from the initial pressure of SbCl5 and add x to both the initial pressures of SbCl3 and Cl2.
Now we can write the expression for Kp:
Kp = (P(SbCl3) * P(Cl2)) / P(SbCl5)
Kp = ((Initial pressure of SbCl3 + x) * (Initial pressure of Cl2 + x)) / (Initial pressure of SbCl5 - x)
Similarly, we can write the expression for Kc:
Kc = ([SbCl3] * [Cl2]) / [SbCl5]
Since the gases are in a closed system, we can assume that the initial concentrations are the same as the initial pressures. However, to find the equilibrium concentrations, we need to convert the pressures to concentrations. Using the ideal gas law, we can write:
P = nRT / V
Concentration = n / V = P / (RT)
Plugging in the expressions for pressure into the concentration equations, we can find the equilibrium concentrations.
(c) In order to produce some SbCl5, a 1.00 mole sample of SbCl3 is first placed in an empty 2.00 liter container maintained at a temperature different from 182ºC. At this temperature, Kc equals 0.117. How many moles of Cl2 must be added to this container to reduce the number of moles of SbCl3 to 0.500 mole at equilibrium?
To solve this problem, we need to set up an ICE (Initial-Change-Equilibrium) table. Let's assume x moles of Cl2 is added to the container. The reaction can be represented as:
SbCl3 + Cl2 <=> SbCl5
Initial: 1.00 mol SbCl3, 0 mol SbCl5, 0 mol Cl2
Change: -x (decrease in SbCl3), +x (increase in SbCl5 and Cl2)
Equilibrium: (1.00 - x) mol SbCl3, x mol SbCl5, x mol Cl2
Using the equation Kc = [SbCl5] / [SbCl3][Cl2], we can set up the expression for the equilibrium constant:
Kc = (x) / ((1.00 - x)(x))
We know that Kc is equal to 0.117. Substituting the values into the equation:
0.117 = (x) / ((1.00 - x)(x))
Now we can rearrange and solve this equation for x:
0.117 * (1.00 - x)(x) = x
0.117 - 0.117x = x^2
x^2 + 0.117x - 0.117 = 0
We can solve this quadratic equation using the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
x = (-(0.117) ± √((0.117)^2 - 4(1)(-0.117))) / (2(1))
x = (-0.117 ± √(0.013689 - (-0.468))) / 2
x = (-0.117 ± √(0.482689)) / 2
x ≈ 0.063 or -0.180
Since we're interested in the number of moles of Cl2, we can neglect the negative value. Therefore, approximately 0.063 moles of Cl2 must be added to the container to reduce the number of moles of SbCl3 to 0.500 mole at equilibrium.
1. What is the concentration in moles per liter of SbCl5 in the container before any decomposition occurs?
moles = grams/molar mass.
Solve for moles.
M = moles/L. Solve for M.
2. What is the pressure in atmospheres of SbCl5 in the container before any decomposition occurs?
Use PV = nRT. Don't forget to change T to Kelvin.
b)If the SbCl5 is 29.2 percent decomposed when equilibrium is
established at 182°C, calculate the values for equilibrium constants Kp and Kc, for this decomposition reaction.
Kc=?
Kp=?
M x fraction decomposed = (SbCl3)
M x fraction decomposed = (Cl2)
M x (1.00 - fraction decomposed) = (SbCl5) after decomposition.
Substitute into the Kc expression below:
Kc = (SbCl3)(Cl2)/(SbCl5) = Kc
For Kp, the following:
You know (Cl2), (SbCl3), and (SbCl5) in M.
Multiply each by 15.0 L (M x L = moles) to obtain moles of each, then use PV = nRT to obtain the pressure of each. Substitute into the Kp expression.
Kp = pSbCl2*pCl2/pSbCl5 = ??
c.In order to produce some SbCl5, a 1.00 mole sample of SbCl3 is first
placed in an empty 2.00 liter container maintained at a temperature
different from 182ºC. At this temperature, Kc, equals 0.117. How many moles of Cl2 must be added to this container to reduce the number of
moles of SbCl3 to 0.500 mole at equilibrium? 1mole/2L = 0.5 M
.....SbCl5(g) ==>SbCl3(g) + Cl2(g)
.....0...........0.5M.......0
.................-x........ +x
................0.25M.......y
What is y? 0.5-x = 0.25
Solve for x to obtain y.
Post your work on these if you get stuck.